Multiply the first equation by 5 to get 15 x − 5 y = 55 .
Add the modified equation to the second equation: ( 15 x − 5 y ) + ( 4 x + 5 y ) = 55 + 2 , which simplifies to 19 x = 57 .
Solve for x : x = 19 57 = 3 .
Substitute x = 3 into the first equation: 3 ( 3 ) − y = 11 , which gives y = − 2 . The solution is x = 3 , y = − 2 .
Explanation
Understanding the Problem We are given a system of two linear equations with two variables, x and y . Our goal is to find the values of x and y that satisfy both equations simultaneously. The equations are:
Equation 1 3 x − y = 11
Equation 2 4 x + 5 y = 2
Solution Strategy We can solve this system of equations using either substitution or elimination. Let's use the elimination method. Multiply the first equation by 5 to eliminate y .
Multiplying Equation 1 by 5 Multiplying Equation 1 by 5, we get:
Resulting Equation 5 ( 3 x − y ) = 5 ( 11 ) 15 x − 5 y = 55
Adding the Equations Now, add this new equation to the second equation (Equation 2) to eliminate y :
Combining Like Terms ( 15 x − 5 y ) + ( 4 x + 5 y ) = 55 + 2 19 x = 57
Solving for x Now, solve for x :
Value of x x = 19 57 x = 3
Substituting x into Equation 1 Substitute the value of x back into either of the original equations to solve for y . Let's use Equation 1:
Substituting x = 3 3 x − y = 11 3 ( 3 ) − y = 11 9 − y = 11
Solving for y Now, solve for y :
Value of y − y = 11 − 9 − y = 2 y = − 2
Final Answer Therefore, the solution to the system of equations is x = 3 and y = − 2 .
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling supply and demand in economics. For example, a company might use a system of equations to determine how many units of a product they need to sell to cover their costs and start making a profit. Understanding how to solve these systems allows for informed decision-making in many practical situations.